Talk:S-expression

I prefer a much bruter method to show the semiparticulars' superparticularity

Demonstration of semiparticulars' superparticularity

If k = 4n: [math]\displaystyle{ \begin {align} {\rm S}k/{\rm S}(k + 2) &= {\rm S}(4n)/{\rm S}(4n + 2) \\ &= \frac {((4n) + 3)(4n)^2}{((4n) - 1)((4n) + 2)^2 } \\ &= \frac {(4n + 3)(4n)^2}{(4n - 1)(4n + 2)^2} \\ &= \frac {(4n + 3)(2n)^2}{(4n - 1)(2n + 1)^2} \\ &= \frac {16n^3 + 12n^2}{16n^3 + 12n^2 - 1} \end {align} }[/math]	If k = 4n + 1: [math]\displaystyle{ \begin {align} {\rm S}k/{\rm S}(k + 2) &= {\rm S}(4n + 1)/{\rm S}(4n + 3) \\ &= \frac {((4n + 1) + 3)(4n + 1)^2}{((4n + 1) - 1)((4n + 1) + 2)^2 } \\ &= \frac {(4n + 4)(4n + 1)^2}{4n(4n + 3)^2} \\ &= \frac {(n + 1)(4n + 1)^2}{n(4n + 3)^2} \\ &= \frac {16n^3 + 24n^2 + 9n + 1}{16n^3 + 24n^2 + 9n} \end {align} }[/math]
If k = 4n - 1: [math]\displaystyle{ \begin {align} {\rm S}k/{\rm S}(k + 2) &= {\rm S}(4n - 1)/{\rm S}(4n + 1) \\ &= \frac {((4n - 1) + 3)(4n - 1)^2}{((4n - 1) - 1)((4n - 1) + 2)^2 } \\ &= \frac {(4n + 2)(4n - 1)^2}{(4n - 2)(4n + 1)^2} \\ &= \frac {(2n + 1)(4n - 1)^2}{(2n - 1)(4n + 1)^2} \\ &= \frac {32n^3 - 6n + 1}{32n^3 - 6n - 1} \end {align} }[/math]	If k = 4n - 2: [math]\displaystyle{ \begin {align} {\rm S}k/{\rm S}(k + 2) &= {\rm S}(4n - 2)/{\rm S}(4n) \\ &= \frac {((4n - 2) + 3)(4n - 2)^2}{((4n - 2) - 1)((4n - 2) + 2)^2 } \\ &= \frac {(4n + 1)(4n - 2)^2}{(4n - 3)(4n)^2} \\ &= \frac {(4n + 1)(2n - 1)^2}{(4n - 3)(2n)^2} \\ &= \frac {16n^3 - 12n^2 + 1}{16n^3 - 12n^2} \end {align} }[/math]

So we see for k = 4n, 4n + 1, and 4n - 2, a coefficient of 2² is canceled out, whereas for the k = 4n - 1, a coefficient of 2 is canceled out. FloraC (talk) 18:31, 7 February 2022 (UTC)

I figured something like this would be possible but I sometimes get lost in how to simplify and group stuff in intermediate steps when I do it that way or I make mistakes simplifying/expanding so I tried to use the most intuitive approach I could think of. The observation that k=2n leads to a factor of 4 I think is a relatively intuitive explanation of why it's superparticular for those cases. Also, it took me a little while but I believe S(k)/S(k+2) = (k+3)/(k-1) * k²/(k+2)² is the equation you substituted into for the four cases? (Just arranged as one single fraction.) --Godtone (talk) 23:45, 7 February 2022 (UTC)

Yup that's the equation I started from. We may go even more primitive if we want, such as from the definitions. It doesn't matter in the end. I consider the elementary algebraic operations an easy problem, i.e. the solution is guaranteed by a known routine. All the tricks on the other hand involve observation and intuition. I mean, it's totally reasonable for one to apply them to instantly gain the insight, but I take the liberty of assuming the readers' expectation here is a solid and sound presentation of the result. FloraC (talk) 00:55, 8 February 2022 (UTC)

What is So supposed to stand for?

[math]\displaystyle{ So(k) = \frac{ 4k^2 + 4k + 1 }{ 4k^2 - 4k - 3 } }[/math] simplifies to [math]\displaystyle{ \frac{2 k + 1}{2 k - 3} }[/math] so it doesn't feel very "square". --Frostburn (talk) 12:08, 5 July 2024 (UTC)

From Square superparticular#Abstraction: "A suggestion is to use the notation Sok, if this is not unambiguous, with the letter "o" standing for "odd"."

Specifically, it's "square" w.r.t. the "odd harmonic series" (as opposed to the natural harmonic series); with respect to odds it's technically only one off of the odd analogue of superparticular.

If that still doesn't make sense to you, think of [math]\displaystyle{ {\rm So}(k) = h_k^2 h_{k-1}^{-1} h_{k+1}^{-1} }[/math] as the "source" of its squareness (as h_k is the analogue of the kth harmonic so this is the direct analogue of k²/(k+1)/(k-1)). Ideally, for the analogy to be perfect, Sok would be odd-particulars (as it'd be impossible to have them be superparticular when they are defined entirely in terms of ratios between odd numbers), but quodd-particular is thus the next best thing.

Hope that clarifies. --Godtone (talk) 22:31, 5 July 2024 (UTC)

I should note that "S" stands for "(Shorthand for) Second-order/Square Superparticular", so by analogy and contrast:

you can think of "So" as standing for ""(Shorthand for) Second-order/Square Odd-particular". --Godtone (talk) 13:17, 6 July 2024 (UTC)

No monzos?

Why --Eufalesio (talk) 15:04, 26 May 2026 (UTC)

There used to be a section on what I call "S-factorizations" (formerly S-monzos before realising that term was taken), but it was moved to S-expression/Advanced results, but S-factorizations are mentioned on the page so probably it should be moved back onto the page, leaving the mathematical derivations section and the abstraction section there. You might ask why not use normal monzos and the answer is it's clearer for the way S-commas work, and this sense is illuminated if you understand the abstraction section whereby you can understand how each infinite comma family of an S-expression is actually readable in a generalised sense based on an abstract harmonic series. But probably that should stay there with the mathematical derivations at least as long as I've written it in terms of group theory. But it does give an additional answer to the question (which also isn't useless as the two examples of potentially interest show), so I wonder if including it might be acceptable under some collapsable section? (IDK if those are possible on MediaWiki...) --Godtone (talk) 23:02, 26 May 2026 (UTC)